Here's a fun way to solve this problem.
Take a pack of playing cards or a set of tiles marked with the digits 0-9. If you use playing cards, the number 10 can represent zero. We'll call the cards or tiles tokens.
Prepare a play area to place the tokens like the lines in the given sum: two rows of three and one row of four.
In the first place on the left of the third row of the play area place 1. This is the thousands place of the sum of the first two rows.
Arrange the tokens in order in your hand with the smallest (0) on top followed by 2-9. when you're holding the tokens always have them in order with smallest at the top.
When placing the tokens in their places make sure that the ones token on the first row has a lower value than the ones token on the second row. And the same for the tens and hundreds tokens.
Now the game starts. Place 2 at the end of the first row (ones place) and under it place 3 and, since 2+3=5, place 5 in the ones place in the third (answer) row. You've now used 4 tokens, so you have 6 left in your hand. Make sure they're in order: 0, 4, 6, 7, 8, 9.
Place the smallest value token (0) in the tens place on the first row. You'll realise that whatever token you place in the tens place on the second row, you would need the same token in the third row, so pick up the 0 and place it at the bottom of the pack you are holding. This means you can't use the 0, so take the token that's now on the top of the pack, which should be 4. Place 4 in the tens place on the first row, then place the next token (6) in the tens place on the second row. 4+6=10 so 0 can now be placed in the tens place of the answer row. Remember that we are carrying 1 in the hundreds place.
You only have 3 tokens in hand (7, 8, 9). If you place 7 in the hundreds place on the first row and 8 in the same place on the second row, you would need 6 for the answer row, but 6 is already in the play area. Why 6 since 7+8=15? Because you have to add in the carry from the tens. Replace the 8 with the 9, and this time you have the sum 7+9=16 plus the carry=17, but 7 is already in the play area. Replace 7 with 8 and you get 8+9=17. And you have 7 in your hand, but, wait, you have to add the carry so you would need 8, which is already in the play area.
Leave 1, 2, 3, 4 and 5 where they are, and pick up the tokens and put them in order again in your hand. You already tried 6 in the tens place of the second row, so the next token to try would be 7, but you would need 1 in the third row because 4+7=11, and 1 is still in the play area. You can't replace 7 with 8, because 4+8=12 and 2 is already in the play area. 4+9=13, but 3 has already been used. So you've tried everything with 4, and you haven't found a solution. So after 4 you need to try 6 in the tens place on row 1 then each of the tokens bigger than 6 in row 2.
That's the way you play the game. Once you get a feel for it you will be able to work faster and see what doesn't work. It may mean starting from scratch again. So, if 2+3=5 doesn't work then you try 2+4=6, and so on. Make a note of what doesn't work. Eventually you will find a solution.
Here's some solutions (there are always 8 solutions related to the same vertical pair of digits):
342+ 346+ 352+ 356+ 742+ 746+ 752+ 756+
756 752 746 742 356 352 346 342
1098 1098 1098 1098 1098 1098 1098 1098
The following solutions only show one arrangement for the same vertical digit pairs, so there will always be 7 more arrangements.
324+ 432+
765 657
1089 1089
Note that the first of these alternative solutions simply transposes two columns of the original solution.
Here's a logical mathematical way to solve the problem. The first digit (thousands place) in the answer can only be 1. The digit 0 must be in either:
(a) the second digit (hundreds place) in the answer, because 0 cannot be the first digit (hundreds place) and it cannot be the last digit (ones place) in either of the two addends; or,
(b) the answer in the tens or ones place. 2+8=3+7=4+6=10. None of these combinations allows the remaining digits to be used to produce the correct sum, because of the presence of the carryover.
Following (a), we isolate 0 and 1 and use these as the first two digits (10) of the 4-digit answer. If then we take the digits 2-9 and note that we can consider the sums that give us 8, 9 or 10 we get:
3+5=2+6=8, 3+6=2+7=4+5=9; 3+7=4+6=10. This ensures that there is no carry into the hundreds place in the answer.
From this we see that we can only use 3+7 and 4+6 to get 10 (the first 2 digits of the answer).
(1) 3+7=10; this leaves us with 2+6=8 (because 3+5 reuses 3) and 4+5=9 (two remaining digits).
This gives us several possibilities: 324+765=1089, 724+365=1089, 742+356=1098, etc. 16 solutions.
(2) 4+6=10; this leaves us with 3+5=8 and 2+7=9:
432+657=1089, 475+623=1098, etc. 16 solutions.
So there appear to be 32 solutions in all, and the 4-digit answer is either 1089 or 1098.